In the figure, LDEH is a trapezium and triangles LHG and GLF are isosceles triangles. KC, KE and CE are straight lines. LH = LG = GF. Find
- ∠v
- ∠w
(a)
∠DLF = 180° - ∠v (Interior angles, LD//KE)
∠GFL = 180° - ∠v (Angles on a straight line)
∠GLF = 180° - ∠v (Isosceles triangle)
58° + 180° - ∠v + 180° - ∠v + ∠w + 23° = 180° (Angles on a straight line, CK)
58° + 180° + 180° + 23° - ∠v - ∠v + ∠w = 180°
441° - 2∠v + ∠w = 180°
2∠v - ∠w = 441° - 180°
2∠v - ∠w = 261°
∠w = 2∠v - 261° --- (1)
∠LGH
= ∠LHG
= 2 x (180° - ∠v)
= 360° - 2∠v (Exterior angle of a triangle)
∠w = 180° - (360° - 2∠v) - (360° - 2∠v)
∠w = 180° - 360° + 2∠v - 360° + 2∠v
∠w = 180° - 360° - 360° + 2∠v + 2∠v
∠w = 4∠v - 540° (Angles sum of triangle)
∠w = 4∠v - 540° --- (2)
(2) = (1)
4∠v - 540° = 2∠v - 261°
4∠v - 2∠v= 540° - 261°
2∠v = 279°
∠v
= 279° ÷ 2
= 139.5°
(b)
From (1)
∠w
= 2∠v - 261°
= 279° - 261°
= 18°
Answer(s): (a) 139.5°; (b) 18°